Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Jeff Root on 25 Oct 2005 06:58 Eric Gisse wrote: > He makes the same elementry mistakes over and over. Those are not mistakes. They are delusions which support his fantasy. > Patience is not the solution. I agree. Patience is only a part of the solution. > The tone was set and continues to be set by Henri Bad choice. > who insists on not calling me by my actual name. It hurts me a little every time I see him do that, and I've never really had my funny last name made fun of. > I am not the type of person to take the high road. We can change that. -- Jeff, in Minneapolis
From: The Ghost In The Machine on 25 Oct 2005 07:00 In sci.physics, HW@..(Henri Wilson) <HW@> wrote on Tue, 25 Oct 2005 08:43:33 GMT <bhrrl1pdji5hc7gqal2ka02j3pdnf8mttj(a)4ax.com>: > On Tue, 25 Oct 2005 04:16:31 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> > wrote: > >>"Eric Gisse" <jowr.pi(a)gmail.com> wrote in >>news:1130198692.816264.53140(a)f14g2000cwb.googlegroups.com: >> > >>>> >I agree. OK in SR, bad for BaT. >>>> >>>> why? >>> >>> You don't place bounds on the v in c+v. c' can be smaller than c. >> >>Exactly right, Dr. Gisse. BaT doesn't place bounds on velocity. SR does. >> >>That presents Henri with a problem. >> >> >>>> >> Try rocket propulsion theory....remember the exhaust accelerates >>>> >> too, mainly in the same direction as the pulse. >>>> > >>>> >????. The exhaust accelerates in all directions, the rocket chamber >>>> >confines the accelerated paricles and only allows them to escape in >>>> >the direction opposite to the light pulse you are trying to catch. >>>> >>>> Wrong. >> >>http://www.suzy.co.nz/suzysworld/Factpage.asp?FactSheet=116 >> >>http://www.straightdope.com/mailbag/mrockets.html >> >> >>>> Initially the propellent moves in the opposite direction to the rocket >>>> motion wrt base. >> >>I think you are confusing yourself. The mass of the rocket is moving, as a >>unit, in the direction that the .2 c BaT photon is going. >> >>The exhaust from the rocket is going in the opposite direction, to provide >>the 'action-reaction' push. We don't know where the base is, nor do we >>care. >> >>>> When the rocket speed wrt base equals the propellent speed wrt the >>>> rocket, the propellent speed wrt base becomes zero. >> >>In order for the rocket to accelerate in one direction, the reaction-mass >>of the fuel must be ejected from the rocket at a much higher velocity in >>the other direction. >> >>>> Further acceleration by the rocket sees the propellent also moving in >>>> the rocket direction wrt base. >> >>Acceleration doesn't have eyes, so IT can't see anything. >> >>>> So energy is eventually required to accelerate both rocket and >>>> propellent in the same direction wrt base. >> >>Forget about the base, it could be in any direction by now. >> >>>> As the rocket approaches c wrt base, so does the propellent speed...and >>>> in the same direction. >> >>No, the exhaust must go in the opposite direction from the rocket. > > A good example of the futility of arguing with non physicists. > > Bob, If the propellent is moving at 1000 m/sec wrt the rocket > and the rocket has reached a speed of 1001 m/sec wrt base, > what is the speed of the propellent wrt base? 1/(1-(1000)(1001)/c^2), of course. What else? :-P At these velocities, that translates into 3209839924060063/3209839924024313 m/s exactly or 1.000000000011137627061220767 m/s or 1 + 1.113762706122076699*10^-11 m/s, so we're not exactly talking easily measurable effect here. Of course with that rocket performance one's going to have used up 63% of one's fuel just getting to 1001 m/s. v_f = v_i + v_e * log(M_i / M_f) [Tsiolkovsky, Newtonian variant] where v_i is the initial velocity of the rocket, v_f the final velocity, v_e the exhaust relative to the rocket, M_i the initial rocket mass (with fuel), and M_f the final rocket mass (no fuel). Note that this is Newtonian; there is a relativistic variant but I'd have to hunt for it, and of course this is for straight-line travel. If v_i = 0, one can rewrite this as v_f/v_e = log(M_i / M_f) M_f = M_i/exp(v_f/v_e) = M_i * exp(-v_f/v_e) Assuming M_i = 100000 kg, M_f = 36751.2 kg. Personally, I'd want a fusion rocket. The Ultimate Engine, which got shot down by Uncle Al -- and for good reason, until we get the bugs worked out in fusion drive :-) -- would have had about 25MeV applied to a helium atom, translating into an approximate exhaust velocity of 34500000 m/s or .115 c, if my computations are correct. One can get even higher exhaust velocities if one wants to discard hydrogen or even electron-positron pairs instead. Sans antimatter, subspace, Asimovian hyperjumping, stargates, or Fred Saberhagen's "C+ theory" (whatever that might be, in Fred Saberhagen's Berserker universe) this is about the best we can do. One could also contemplate a deuterium-tritium fusion power source, yielding about 17 MeV plus a free neutron, but the tritium doesn't last long when not in use. > > When the rocket reaches 2900000000 m/sec, what is that speed? > Sorry. Tachyonic rockets are not allowed. :-P Did you mean 290000000 m/s? If so, the exhaust velocity relative to the base is as one might expect: (290000000-1001)/(1-(290000000)(1001)/c^2) = 289999935.6742280716613296097 Scoff as one will, but that is what SR predicts. > > >> >>>> KE approaches 1/2mc^2. Expended energy dissipated as heat must be at >>>> least that. >>>> Total final energy = mc^2 >> >> >> >>> Oh you are using Newtonian kinematics, to prove the validity of the >>> "BaT", while using an example derived from special relativity? >>> >>> Well, that is WRONG, ****head! For so many g*dd*mn reasons. >> >>Please, watch the language. No need for such terms. Leave them to A**r*cl*s >>and his ilk. >> >>> Newtonian mechanics sets c=oo, and SR sets c = constant in all inertial >>> frames. >>> >>> Your analysis is dead on arrival because you are using a theory that is >>> incompatable with the premise of your theory. >>> >>> Your analysis is dead on arrival due to the above plus E = mc^2 is >>> derived from SR and not Newton. I asked for you to show me otherwise >>> and since you are incapable of doing it, my point stands. >>> >>>> >>>> get it? >>>> >>>> I say there can never be enough energy in chemical or nuclear bonds to >>>> get to c...because maximum available is E=mc^2. >> >>I agreed already that in SR universe we can't reach c with matter. >> >>But in BaT universe, c isn't a magic number. Newtonian Relativity says that >>I can keep accelerating, and if you run the numbers, without relativistic >>mass increase as you approach c, you will find that it is easy for the >>rocket to exceed c.... And it would IF we lived in a BaTty universe. > > Bob, there is no mass increase. That is a leftover from aether theory. > > In relativity, contractions aren't real. Maybe not, but the LHC designers don't seem to want to design for superluminal protons. > >>> That is derivable from SR, not Newton. Goddamn you are dumb. >> >>He is just a bit muxed up. >>No need to be rude. Name calling makes the name caller look bad. >> >>>> However, maybe some of the accelerating energy could come from an >>>> external source... >> >>magic? > > No, external boosters are used to launch rocket regularly. External boosters are still part of the rocket, until shed. > >> >>> No sh*t. Perhaps we could test the hypothesis of c'=c+v using that >>> method. >>> >>> Henri just thought of the concept now known as the "particle >>> accelerator". Only a century and change late Henri, good job! > > Geese seems to have discovered drinking. > > Next he might discover that women are better than his fist. That's too much information. :-P (FSVO "information".) [.sigsnip] -- #191, ewill3(a)earthlink.net It's still legal to go .sigless.
From: Paul B. Andersen on 25 Oct 2005 07:40 Henri Wilson wrote: > On Mon, 24 Oct 2005 14:17:05 +0200, "Paul B. Andersen" > <paul.b.andersen(a)deletethishia.no> wrote: > > >>HW@.. wrote: >> >>>On Sun, 23 Oct 2005 15:46:51 +0200, "Paul B. Andersen" >>><paul.b.andersen(a)deletethishia.no> wrote: >>> > > >>>>>It does. It must. >>>> >>>>Ok. >>>>So you claim that the BaT predicts what the light curve >>>>should look like. >>>> >>>> >>>> >>>>>>If it does, does it then predict this light curve: >>>>>>http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1978MNRAS.184..523N&data_type=PDF_HIGH&type=PRINTER&filetype=.pdf >>>>>>or >>>>>>http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1978MNRAS.184..523N& >>>>>>And retrieve the full article. >>>>>> >>>>>>If it does not, what does it then predict? >>>>>>Nothing? >>>> >>>>The last time I showed you this, >>>>Henri Wilson wrote: >>>>| It is willusory by definition. >>>>| Because light is used for gaining information about the star, it is a >>>>| willusion. >>>>| The task is to find te truth that causes the willusion. >>>> >>>> >>>> >>>>>If it does not, it is simply because not all factirs have yet been considered >>>>>(in my program anyway) >>>> >>>>It does not. >>>>The factor you haven't considered is that >>>>the BaT is wrong. >>> >>> >>>I used to consider that..but soon realised that it cannot be wrong. >> >>The point is that you now have admitted that the BaT >>predicts what the light curve should look like, despite >>the fact that you called this light curve a "willusion". >> >>But according to you, the BaT predicts a light curve >>quite different from this: >>http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1978MNRAS.184..523N&data_type=PDF_HIGH&type=PRINTER&filetype=.pdf >>or >>http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1978MNRAS.184..523N& >> And retrieve the full article. >> >>Wrong prediction -> theory falsified. >>BaT is falsified. > > > There are other factors to consider. Such as? > >>>>>>>Different layers within the star have different radial velocities...and that >>>>>>>includes gaseous layers far beyond the extremities of the main body. >>>>>> >>>>>>This is caused by the rotation _only_. >>>>>>The orbital motion has nothing to do with it. >>>>> >>>>> >>>>>Run my little program: >>>>>http://www.users.bigpond.com/hewn/radialvs.exe >>>>> >>>>>(select 'star rotation =1' for tidal lock) >>>> >>>>So the star is rotating. >>> >>> >>>Most - ney ALL - stars rotate. >> >>Indeed they do. >> >> >>>Are you able to run my program? It takes only about five seconds to download. >>>You will enjoy it. >>>A picture speaks a thousand words. >> >>I do not need a picture to understand that >>different parts of a rotating star have different >>radial velocities. > > > That is not the point. You seem to want to avoid the important point. > > .....which is that in a rotating star, the average radial velocity (wrt a > distant observer) of all elements of a particular spherical layer is not the > same as that of another layer. I didn't avoid that point. I said why it is irrelevent. All the radiation comes from the same layer, the photosphere. >>>>>Depends how hot it is and how fast it is rotating. >>>> >>>>In Wondersland, yes. >>>>But not in the real world. >>> >>> >>>Can you not see that the light from the edges would be doppler shifted both >>>ways. It the star was rotating fast enough that would broaden the radiation >>>curve away from precise black body. >> >>You have no sense of proportions, have you? >>Of course a rotating star will broaden the black body >>spectrum, but only to such a small degree that it >>never will be detectable. >> >>As you so correctly said above, all stars are rotating. >>And all stars radiates a black body spectrum. >>The broadening of the BB spectrum due to the stars rotation >>is never observable. > > > OK you accept that it happens. So do you accept that this effect is way too small to be observed? >>>>>run my little program: >>>>>http://www.users.bigpond.com/hewn/radialvs.exe >>>>> >>>>>If you cannot see that the green dots have a higher radial velocities wrt the >>>>>observer than the yellow ones then there is something wrong with you. >>>> >>>>Of course, the star is rotating. >>>>Just about _all_ stars are rotating. >>>>So what? >>> >>> >>>You still haven't run the program. >> >>I have. >>Different parts of the star have different >>radial velocity _because the star is rotating_. >>So what? > > > I DON'T BELIEVE YOU RAN IT. > You refuse to accept the truth. What the hell are you fussing about? >>>>The secondary minimum of Algol is practically unobservable >>>>in visible light, while it is 0.35 magnitudes deep at 10um, >>>>exactly as the conventional theory predicts they should be. >>>>The BaT does _not_ predict that the light curve >>>>in 10um and visible light should be different. >>> >>> >>>Bull. >> >>I think you know that the BaT predicts the same >>light curve for 10um as for visible light. > > > there are several factors to be considered apart from the one I gave. And the factor you gave was that different parts of the spectrum are emitted from different depths of the star. That is wrong, so you have still to mention a single factor. So what are the factors to be considered? The truth is simple, Henri. The BaT is falsified. Again. Paul
From: bz on 25 Oct 2005 08:04 "Eric Gisse" <jowr.pi(a)gmail.com> wrote in news:1130227764.910945.54140(a)g49g2000cwa.googlegroups.com: > > bz wrote: > > [snip] > >> >> > >> >> >I agree. OK in SR, bad for BaT. >> >> >> >> why? >> > >> > You don't place bounds on the v in c+v. c' can be smaller than c. >> >> Exactly right, Dr. Gisse. BaT doesn't place bounds on velocity. SR >> does. > > I don't have a doctorate, or even a bachelor's. I work daily with PhDs, Post Docs, Grad Students, undergrads and never grads. Intellegence and ability have little correlation with degrees obtained. The degrees show an ability to stick to the task, however. > I am, however, a bad > student. At the very least I am on par with Henri, because I too can be > very loud about my ignorance. You often make cogent comments. I enjoy reading those. >> >> > Oh you are using Newtonian kinematics, to prove the validity of the >> > "BaT", while using an example derived from special relativity? >> > >> > Well, that is WRONG, ****head! For so many g*dd*mn reasons. >> >> Please, watch the language. No need for such terms. Leave them to >> A**r*cl*s and his ilk. > > He makes the same elementry mistakes over and over. Patience is not the > solution. Nothing else will work better. > The tone was set and continues to be set by Henri who insists on not > calling me by my actual name. I have advised him, numerous times, via e-mail, that he only makes himself look bad by such actions. > I am not the type of person to take the high road. There is only one person that you can control. You. Your actions DO have an effect on others, however. Taking the high road costs less in the long run. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: bz on 25 Oct 2005 08:26
HW@..(Henri Wilson) wrote in news:bhrrl1pdji5hc7gqal2ka02j3pdnf8mttj(a)4ax.com: > On Tue, 25 Oct 2005 04:16:31 +0000 (UTC), bz > <bz+sp(a)ch100-5.chem.lsu.edu> wrote: > >>"Eric Gisse" <jowr.pi(a)gmail.com> wrote in >>news:1130198692.816264.53140(a)f14g2000cwb.googlegroups.com: >> > >>>> >I agree. OK in SR, bad for BaT. >>>> >>>> why? >>> >>> You don't place bounds on the v in c+v. c' can be smaller than c. >> >>Exactly right, Dr. Gisse. BaT doesn't place bounds on velocity. SR does. >> >>That presents Henri with a problem. >> >> >>>> >> Try rocket propulsion theory....remember the exhaust accelerates >>>> >> too, mainly in the same direction as the pulse. >>>> > >>>> >????. The exhaust accelerates in all directions, the rocket chamber >>>> >confines the accelerated paricles and only allows them to escape in >>>> >the direction opposite to the light pulse you are trying to catch. >>>> >>>> Wrong. >> >>http://www.suzy.co.nz/suzysworld/Factpage.asp?FactSheet=116 >> >>http://www.straightdope.com/mailbag/mrockets.html >> >> >>>> Initially the propellent moves in the opposite direction to the >>>> rocket motion wrt base. >> >>I think you are confusing yourself. The mass of the rocket is moving, as >>a unit, in the direction that the .2 c BaT photon is going. >> >>The exhaust from the rocket is going in the opposite direction, to >>provide the 'action-reaction' push. We don't know where the base is, nor >>do we care. >> >>>> When the rocket speed wrt base equals the propellent speed wrt the >>>> rocket, the propellent speed wrt base becomes zero. >> >>In order for the rocket to accelerate in one direction, the >>reaction-mass of the fuel must be ejected from the rocket at a much >>higher velocity in the other direction. >> >>>> Further acceleration by the rocket sees the propellent also moving in >>>> the rocket direction wrt base. >> >>Acceleration doesn't have eyes, so IT can't see anything. >> >>>> So energy is eventually required to accelerate both rocket and >>>> propellent in the same direction wrt base. >> >>Forget about the base, it could be in any direction by now. >> >>>> As the rocket approaches c wrt base, so does the propellent >>>> speed...and in the same direction. >> >>No, the exhaust must go in the opposite direction from the rocket. > > A good example of the futility of arguing with non physicists. Henri, there are many here that understand physics better than you or I do. I understand that the delta KE of fuel the expelled = delta KE of rocket in any instance of acceleration. > > Bob, If the propellent is moving at 1000 m/sec wrt the rocket and the > rocket has reached a speed of 1001 m/sec wrt base, what is the speed of > the propellent wrt base? How does the speed wrt the base have anything to do with catching up to a photon that was fired from an object moving away? under BaT photon<------ <<rocket<<--- -->>>laser that fired photon>>> laser is moving at .9c that ---> way. photon is moving at .1c <---- that way. Rocket is moving at .2c <---- that way. It will catch the photon. everything is measured wrt the original rest frame of the rocket, what you are calling 'base'. -------------------------------------------------------------- > When the rocket reaches 2900000000 m/sec, what is that speed? 9.673 c >>>> KE approaches 1/2mc^2. Expended energy dissipated as heat must be at >>>> least that. >>>> Total final ..... >>> Newtonian mechanics sets c=oo, and SR sets c = constant in all >>> inertial frames. >>> >>> Your analysis is dead on arrival because you are using a theory that >>> is incompatable with the premise of your theory. >>> >>> Your analysis is dead on arrival due to the above plus E = mc^2 is >>> derived from SR and not Newton. I asked for you to show me otherwise >>> and since you are incapable of doing it, my point stands. >>> >>>> >>>> get it? >>>> >>>> I say there can never be enough energy in chemical or nuclear bonds >>>> to get to c...because maximum available is E=mc^2. >> >>I agreed already that in SR universe we can't reach c with matter. >> >>But in BaT universe, c isn't a magic number. Newtonian Relativity says >>that I can keep accelerating, and if you run the numbers, without >>relativistic mass increase as you approach c, you will find that it is >>easy for the rocket to exceed c.... And it would IF we lived in a BaTty >>universe. > > Bob, there is no mass increase. That is a leftover from aether theory. Then how do you limit mass to < c ? > > In relativity, contractions aren't real. Who said they were? ..... >> >>>> However, maybe some of the accelerating energy could come from an >>>> external source... >> >>magic? > > No, external boosters are used to launch rocket regularly. The boosters are part of the rocket until they are shed. Perhaps you are using a railgun or terawatt ground based laser for propulsion? ..... -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap |